Question

the first four terms of an A.P.is 26 .if the sum of squares of first and fourth term is 125, then find the first four terms

Answer 1

First four terms of ap =26

Answer 2

Answer:

The first four terms are 2,5,8,11.

Step-by-step explanation:

Given : The first four terms of an A.P.is 26 . If the sum of squares of first and fourth term is 125.

To find : The first four terms?

Solution :

Let $a_1$ be the 1st term and d be the common difference:

$a_1+(a_1+d)+(a_1+2d)+(a_1+3d) = 26$

$4a_1 + 6d = 26$

$2(2a_1+3d)=26$

$2a_1 + 3d = 13$......[1]

Sum of squares of first and fourth term is 125

$a_1^2+ (a_1+3d)^2 = 125$

$a_1^2 + a_1^2 + 6a_1d + 9d^2 = 125$

$2a_1^2 + 6a_1d + 9d^2 = 125$ ......[2]

From equation [1] : $a_1= \frac{13-3d}{2}$

Plug into equation [2],

$2[\frac{13-3d}{2}]^2 + 6d\times\frac{(13-3d)}{2} + 9d^2 = 125$

$2[\frac{(169-78d+9d^2)}{4}] + 3d(13-3d) + 9d^2 = 125$

$\frac{(169-78d+9d^2)}{2} + 39d - 9d^2 + 9d^2 = 125$

$\frac{(169-78d+9d^2)}{2} + 39d = 125$

$169 - 78d + 9d^2 + 78d = 250$

$9d^2 + 169 = 250$

$9d^2 = 81$

$d^2 = 9$

$d = 3$

Substitute the value of d in equation [1]

$2a_1 + 3d = 13$

$2a_1 + 9 = 13$

$2a_1=4$

$a_1=2$

Now, we find the four terms with a=2 and d=3

$a_2 = 2+3 = 5\\a_3 = 5+3 = 8\\a_4 = 8+3 = 11$

Therefore, The first four terms are 2,5,8,11.